# solving the problem

Question:

If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$, verify that $A^{2}-4 A+I=O$, where $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $O=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$. Hence, find $A^{-1}$.

Solution:

$A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$\therefore A^{2}=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]$

and

$A^{2}-4 A+I=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]-\left[\begin{array}{cc}8 & 12 \\ 4 & 8\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-4 A+I=\left[\begin{array}{cc}7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O$

$\Rightarrow A^{2}-4 A+I=0$

$\Rightarrow A^{2}-4 A=-I$

$\Rightarrow A^{-1} A^{2}-4 A A^{-1}=-I A^{-1}$      $\left[\right.$ Pre $-$ multiplying both sides by $\left.A^{-1}\right]$

$\Rightarrow A-4 I=-A^{-1}$

$\Rightarrow A^{-1}=4 I-A$

$\Rightarrow A^{-1}=\left\{\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]-\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\right\}=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$